Semisimple algebra

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In ring theory, a semisimple algebra is an associative algebra which has trivial Jacobson radical (that is only the zero element of the algebra is in the Jacobson radical). If the algebra is finite dimensional this is equivalent to saying that it that can be expressed as a Cartesian product of simple subalgebras.

[edit] Definition

Given an algebra, its radical is the (unique) nilpotent ideal that contains all nilpotent ideals in the algebra. A finite dimensional algebra is then said to be semi-simple if its radical is {0}, where 0 denotes the zero element of the algebra.

A algebra A is called simple if it has no proper ideals and A² = {ab | a, b ∈ A} ≠ {0}. As the terminology implies, simple algebras are semi-simple. Only possible ideals in a simple algebra are A and {0}. Thus if A is not nilpotent, then A is semisimple. Because A² is an ideal of A and A is simple, A² = A. By induction, Aⁿ = A for every positive integer n, i.e. A is not nilpotent.

Any self-adjoint subalgebra A of n × n matrices with complex entries is semisimple. Let Rad(A) be the radical of A. Suppose a matrix M is in Rad(A). Then M*M lies in some nilpotent ideals of A, therefore (M*M)^k = 0 for some positive integer k. By positive-semidefiniteness of M*M, this implies M*M = 0. So M x is the zero vector for all x, i.e. M = 0.

If {A_i} is a finite collection of simple algebras, then their Cartesian product ∏ A_i is semi-simple. If (a_i) is an element of Rad(A). Let e₁ be the multiplicative identity in A₁ (all simple algebras possess a multiplicative identity). Then (a₁, a₂, ...) · (e₁, 0, ...) = (a₁, 0..., 0) lies in some nilpotent ideal of ∏ A_i. This implies, for all b in A₁, a₁b is nilpotent in A₁, i.e. a₁ ∈ Rad(A₁). So a₁ = 0. Similarly, a_i = 0 for all other i.

It is less apparent from the definition that the converse of the above is also true, that is, any semisimple algebra is isomorphic to a Cartesian product of simple algebras. The following is a semisimple algebra that appears not to be of this form. Let A be an algebra with Rad(A) ≠ A. The quotient algebra B = A ⁄ Rad(A) is semisimple: If J is a nonzero nilpotent ideal in B, then its preimage under the natural projection map is a nilpotent ideal in A which is strictly larger than Rad(A), a contradiction.

[edit] Characterization

Let A be a finite dimensional semisimple algebra, and

be a composition series of A, then A is isomporphic to the following Cartesian product:

where each

is a simple algebra.

The proof can be sketched as follows. First, invoking the assumption that A is semisimple, one can show that the J₁ is a simple algebra (therefore unital). So J₁ is a unital subalgebra and an ideal of J₂. Therefore one can decompose

By maximality of J₁ as an ideal in J₂ and also the semisimplicity of A, the algebra

is simple. Proceed by induction in similar fashion proves the claim. For example, J₃ is the Cartesian product of simple algebras

The above result can be restated in a different way. For a semisimple algebra A = A₁ ×...× A_n expressed in terms of its simple factors, consider the units e_i ∈ A_i. The elements E_i = (0,...,e_i,...,0) are idempotents in A and they lie in the center of A. Furthermore, E_i A = A_i, E_iE_j = 0 for i ≠ j, and Σ E_i = 1, the multiplicative identity in A.

Therefore, for every semisimple algebra A, there exists idempotents {E_i} in the center of A, such that

1. E_iE_j = 0 for i ≠ j (such a set of idempotents is called orthogonal),
2. Σ E_i = 1,
3. A is isomorphic to the Cartesian product of simple algebras E₁ A ×...× E_n A.

[edit] Classification

The Artin–Wedderburn theorem completely classifies semisimple algebras: they are isomorphic to a product where the n_i are some integers, the D_i are division rings, and means the ring of matrices over D_i. This product is unique up to permutation of the factors.